INTEGRATION BY SUBSTITUTION
(In this section we will study a technique, called substitution, that can often be used
to transform complicated integration problems into simpler ones).
U-SUBSTITUTION
The method of substitution can be motivated by examining the chain rule from the
viewpoint of antidifferentiation. For this purpose, suppose that 𝐹 is an antiderivative
of 𝑓 and that 𝑔 is a differentiable function. The chain rule implies that the derivative
of 𝐹(𝑔(𝑥 )) can be expressed as
𝑑
𝑑𝑥
[𝐹(𝑔(𝑥 ))] = 𝐹′(𝑔(𝑥)) ∙ 𝑔′(𝑥)
which we can write in integral form as
𝑑
∫ 𝐹 ′ (𝑔(𝑥 )) ∙ 𝑔′ (𝑥 )𝑑𝑥 = ∫ 𝑑𝑥 [𝐹(𝑔(𝑥 ))]𝑑𝑥 = 𝐹(𝑔(𝑥 )) + 𝐶
(1)
or since 𝐹 is an antiderivative of 𝑓,
∫ 𝑓(𝑔(𝑥 )) ∙ 𝑔′ (𝑥 )𝑑𝑥 = 𝐹(𝑔(𝑥 )) + 𝐶 .
For our purposes it will be used to let 𝑢 = 𝑔(𝑥) and to write
(2)
𝑑𝑢
𝑑𝑥
= 𝑔′(𝑥) in the
differential form 𝑑𝑢 = 𝑔′ (𝑥 )𝑑𝑥 . With this notation (2) can be expressed as
∫ 𝑓(𝑔(𝑥 )) ∙ 𝑔′ (𝑥 )𝑑𝑥 = ∫ 𝑓(𝑔(𝑥 )) ∙ 𝑑[𝑔(𝑥 )] = ∫ 𝑓(𝑢) ∙ 𝑑𝑢 = 𝐹 (𝑢) + 𝐶
(3)
The process of evaluating an integral of form (2) by converting it into form (3) with
the substitution 𝑢 = 𝑔(𝑥) and 𝑑𝑢 = 𝑔′ (𝑥 )𝑑𝑥 is called the method of u-substitution.
Here the differential serves primarily as a useful “bookkeeping” device for the
method of u-substitution. The following example illustrates how the method works.
Example 1 Evaluate ∫(𝑥 2 + 1)50 ∙ 2𝑥𝑑𝑥 .
Solution
Example 2 1) Evaluate ∫ sin(𝑥 + 9) 𝑑𝑥
2) Evaluate ∫(𝑥 − 8)23 𝑑𝑥
Example 3 Evaluate ∫ cos 5𝑥 𝑑𝑥
Example 4 Evaluate ∫
𝑑𝑥
5
1
(3𝑥−8)
𝑑𝑥
Example 5 Evaluate ∫
1+3𝑥 2
1
Example 6 Evaluate ∫ ( + 𝑠𝑒𝑐 2 𝜋𝑥) 𝑑𝑥
𝑥
Example 7 Evaluate ∫ 𝑠𝑖𝑛2 𝑥 ∙ cos 𝑥 𝑑𝑥
Example 8 Evaluate ∫
𝑒 √𝑥
√𝑥
𝑑𝑥
3
Example 9 Evaluate ∫ 𝑡 4 √3 − 5𝑡 5 𝑑𝑡
𝑒𝑥
Example 10 Evaluate ∫
𝑑𝑥
√1−𝑒 2𝑥
H.W. p. 273 # 1 – 30